=E Δ l l o. π d o 2 4. Δ l = 4Fl o π d o 2 E. = 0.50 mm (0.02 in.)

Transcription

1 6.10 (a) This portion of the problem asks that the tangent modulus be determined for the gray cast iron, the stress-strain behavior of which is shown in Figure The slope (i.e., σ/ ε) of a tangent drawn through this curve at 25 MPa (3625 psi) is about 90 GPa (15 x 10 6 psi). (b) The secant modulus taken from the origin is calculated by taking the slope of a secant drawn from the origin through the stress-strain curve at 35 MPa (5,000 psi). When such a secant is drawn, a modulus of approximately 100 GPa (14.5 x 10 6 psi) is obtained (a) We are asked, in this portion of the problem, to determine the elongation of a cylindrical specimen of aluminum. Using Equations (6.1), (6.2), and (6.5) F π d o 2 4 E Δ l l o Or Δ l 4Fl o π d o 2 E (4)(48,800 N) ( 200 x 10 3 m) ( ) 2 ( 69 x 10 9 N/ m 2 ) (π) 19x10 3 m 0.50 mm (0.02 in.) (b) We are now called upon to determine the change in diameter, Δd. Using Equation (6.8) ν ε x ε z Δd/d o Δl/ l o From Table 6.1, for Al, ν Now, solving for d yields Δd νδld o l o (0.33)(0.50 mm)(19 mm) 200 mm x 10-2 mm (-6.2 x 10-4 in.)

2 The diameter will decrease (a) This part of the problem asks that we ascertain which of the metals in Table 6.1 experience an elongation of less than mm when subjected to a stress of 28 MPa. The maximum strain that may be sustained is just ε Δl mm 3.2 x 10-4 l o 250 mm Since the stress level is given, using Equation (6.5) it is possible to compute the minimum modulus of elasticity which is required to yield this minimum strain. Hence E σ ε 28 MPa 87.5 GPa 3.2 x 10 4 Which means that those metals with moduli of elasticity greater than this value are acceptable candidates--namely, brass, Cu, Ni, steel, Ti and W. (b) This portion of the problem further stipulates that the maximum permissible diameter decrease is 1.2 x 10-3 mm. Thus, the maximum possible lateral strain ε x is just ε x Δd d o 1.2 x 10 3 mm 12.7 mm 9.45 x 10-5 Since we now have maximum permissible values for both axial and lateral strains, it is possible to determine the maximum allowable value for Poisson's ratio using Equation (6.8). Thus ν ε x ε z 9.45 x x Or, the value of Poisson's ratio must be less than Of the metals in Table 6.1, only W meets both of these criteria This problem asks that we assess the four alloys relative to the two criteria presented. The first criterion is that the material not experience plastic deformation when the tensile

3 load of 35,000 N is applied; this means that the stress corresponding to this load not exceed the yield strength of the material. Upon computing the stress σ F A o π F d 2 35, 000 N o π 15 x x 106 N/m MPa m 2 2 Of the alloys listed in the table, the Al, Ti and steel alloys have yield strengths greater than 200 MPa. Relative to the second criterion, it is necessary to calculate the change in diameter d for these two alloys. From Equation (6.8) ν ε x ε z Δd/d o σ /E Now, solving for d from this expression, Δd νσ d o E For the aluminum alloy (0.33)(200 MPa)(15 mm) Δd 70 x x 10-3 mm MPa Therefore, the Al alloy is not a candidate. For the steel alloy (0.27)(200 MPa)(15 mm) Δd 205 x 10 3 MPa 0.40 x 10-2 mm Therefore, the steel is a candidate. For the Ti alloy

4 (0.36)(200 MPa)(15 mm) Δd 105 x 10 3 MPa 1.0 x 10-2 mm Hence, the titanium alloy is also a candidate This problem calls for us to make a stress-strain plot for a magnesium, given its tensile load-length data, and then to determine some of its mechanical characteristics. (a) The data are plotted below on two plots: the first corresponds to the entire stressstrain curve, while for the second, the curve extends just beyond the elastic region of deformation.

5 (b) The elastic modulus is the slope in the linear elastic region as E Δσ Δε 50 MPa 0 MPa x 10 3 MPa 50 GPa ( 7.3 x 10 6 psi) (c) For the yield strength, the strain offset line is drawn dashed. It intersects the stress-strain curve at approximately 150 MPa (21,750 psi). (d) The tensile strength is approximately 240 MPa (34,800 psi), corresponding to the maximum stress on the complete stress-strain plot. (e) From Equation (6.14), the modulus of resilience is just U r σ y 2 2E which, using data computed in the problem, yields a value of U r ( 150 x 10 6 N/m 2 ) 2 ( ) 2.25 x 105 J/m 3 ( 32.6 in. - lb f /in. 3 ) (2) 50 x 10 9 N/m 2 (f) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by one-hundred. The total fracture strain at fracture is 0.110; subtracting out the elastic

6 strain (which is about 0.003) leaves a plastic strain of Thus, the ductility is about 10.7%EL We are asked to compute the true strain that results from the application of a true stress of 600 MPa (87,000 psi); other true stress-strain data are also given. It first becomes necessary to solve for n in Equation (6.19). Taking logarithms of this expression and after rearrangement we have n log σ T log K log ε T log (575 MPa) log (860 MPa) log (0.2) Expressing ε T as the dependent variable, and then solving for its value from the data stipulated in the problem, leads to ε T σ T K 1/n 1/ MPa 860 MPa (a) In order to determine the final length of the brass specimen when the load is released, it first becomes necessary to compute the applied stress using Equation (6.1); thus σ F A o F π d 2 11,750 N o π 10 x m MPa (22,000 psi) Upon locating this point on the stress-strain curve (Figure 6.12), we note that it is in the linear, elastic region; therefore, when the load is released the specimen will return to its original length of 120 mm (4.72 in.). (b) In this portion of the problem we are asked to calculate the final length, after load release, when the load is increased to 23,500 N (5280 lb f ). Again, computing the stress

7 23, 500 N σ π 10 x m MPa (44,200 psi) The point on the stress-strain curve corresponding to this stress is in the plastic region. We are able to estimate the amount of permanent strain by drawing a straight line parallel to the linear elastic region; this line intersects the strain axis at a strain of about which is the amount of plastic strain. The final specimen length l i may be determined from Equation (6.2) as l i l o (1 + ε) (120 mm)( ) mm (4.78 in.) 6.48 (a) We are asked to compute the Brinell hardness for the given indentation. It is necessary to use the equation in Table 6.4 for HB, where P 1000 kg, d 2.50 mm, and D 10 mm. Thus, the Brinell hardness is computed as HB 2P πd D D 2 d 2 (2)(1000 kg) (π)(10 mm) 10 mm (10 mm) 2 (2.50 mm) (b) This part of the problem calls for us to determine the indentation diameter d which will yield a 300 HB when P 500 kg. Solving for d from this equation in Table 6.4 gives d D 2 D 2 2P (HB)πD (10 mm) 2 2 (2)(500 kg) 10 mm (300)(π)(10 mm) 1.45 mm 6.49 This problem calls for estimations of Brinell and Rockwell hardnesses.

8 (a) For the brass specimen, the stress-strain behavior for which is shown in Figure 6.12, the tensile strength is 450 MPa (65,000 psi). From Figure 6.19, the hardness for brass corresponding to this tensile strength is about 125 HB or 70 HRB. (b) The steel alloy (Figure 6.24) has a tensile strength of about 1970 MPa (285,000 psi). This corresponds to a hardness of about 560 HB or ~55 HRC from the line (extended) for steels in Figure We are asked to compute the critical resolved shear stress for Zn. As stipulated in the problem, φ 65, while possible values for λ are 30, 48, and 78. (a) Slip will occur along that direction for which (cos φ cos λ) is a maximum, or, in this case, for the largest cos λ. The cosines for the possible λ values are given below. cos(30 ) 0.87 cos(48 ) 0.67 cos(78 ) 0.21 Thus, the slip direction is at an angle of 30 with the tensile axis. (b) From Equation (7.3), the critical resolved shear stress is just τ crss σ y (cos φ cos λ) max (2.5 MPa) [ cos(65 )cos(30 ) ] 0.90 MPa (130 psi) 7.14 This problem asks that, for a metal that has the FCC crystal structure, we compute the applied stress(s) that are required to cause slip to occur on a (111) plane in each of the [1 1 0 ], [10 1 ], and [01 1] directions. In order to solve this problem it is necessary to employ Equation (7.3), which means that we will need to solve for the for angles λ and φ for the three slip systems. In the sketch below is shown the unit cell and (111)-[1 1 0 ] slip configuration.

9 The angle between the [100] and [1 1 0] directions, λ, is 45. For the (111) plane, the angle between its normal (which is the [111] direction) and the [100] direction, φ, is tan -1 a , therefore, solving for the yield strength from Equation (7.3) a σ y τ crss (cosφ cosλ) 0.5 MPa cos (54.7 ) cos(45 ) 0.5 MPa 1.22 MPa (0.578)(0.707) The (111)- [10 1 ] slip configuration is represented in the following sketch.

10 For this situation values for λ and φ are the same as for the previous situation i.e., λ 45 and φ This means that the yield strength for this slip system is the same as for (111)-[1 1 0 ]; that is σ y 1.22 MPa. The unit cell and final (111)- [01 1] slip system are presented in the following illustration. In this case, φ has the same value as previously (i.e., 54.7 ); however the value for λ is 90. Thus, the yield strength for this configuration is σ y τ crss (cosφ cosλ) 0.5 MPa cos (54.7 ) cos(90 ) 0.5 MPa (0.578)(0) which means that slip will not occur on this (111)- [01 1] slip system Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.

11 7.27 In order for these two cylindrical specimens to have the same deformed hardness, they must be deformed to the same percent cold work. For the first specimen %CW A o A d A o x 100 π r 2 o 2 πrd πr 2 o x 100 π (15 mm)2 π(12 mm) 2 π (15 mm) 2 x %CW For the second specimen, the deformed radius is computed using the above equation and solving for r d as r d r o 1 %CW 100 (11 mm) 1 36%CW mm 7.28 We are given the original and deformed cross-sectional dimensions for two specimens of the same metal, and are then asked to determine which is the hardest after deformation. The hardest specimen will be the one that has experienced the greatest degree of cold work. Therefore, all we need do is to compute the %CW for each specimen using Equation (7.6). For the circular one %CW A o A d A o x mm 11.4 mm π π mm π 2 x %CW For the rectangular one

12 (125 mm)(175 mm) (75 mm)(200 mm) %CW x %CW (125 mm)(175 mm) Therefore, the deformed circular specimen will be harder 7.D4 This problem asks us to determine which of copper, brass, and a 1040 steel may be coldworked so as to achieve a minimum yield strength of 345 MPa (50,000 psi) while maintaining a minimum ductility of 20%EL. For each of these alloys, the minimum cold work necessary to achieve the yield strength may be determined from Figure 7.17(a), while the maximum possible cold work for the ductility is found in Figure 7.17(c). These data are tabulated below. Yield Strength Ductility (> 345 MPa) (> 20%EL) Steel Any %CW < 5%CW Brass > 20%CW < 23%CW Copper > 54%CW < 15%CW Thus, both the 1040 steel and brass are possible candidates since for these alloys there is an overlap of percents coldwork to give the required minimum yield strength and ductility values. 8.3 This problem asks that we compute the magnitude of the maximum stress that exists at the tip of an internal crack. Equation (8.1) is employed to solve this problem, as 1/ 2 a σ m 2σ o ρ t 2.5 x /2 mm (2)(170 MPa) x MPa (354,000 psi) mm

13 8.4 In order to estimate the theoretical fracture strength of this material it is necessary to calculate σ m using Equation (8.1) given that σ o 1035 MPa, a 0.5 mm, and ρ t 5 x 10-3 mm. Thus, σ m 2σ o a ρ t (2)(1035 MPa) 0.5 mm 5 x10 3 mm 2.07 x 104 MPa 207 GPa ( 3 x 10 6 psi) 8.7 The maximum allowable surface crack length for MgO may be determined using Equation (8.3); taking the value of 225 GPa as the modulus of elasticity (Table 12.5), and solving for a, leads to a 2Eγ (2) s ( 225 x 10 9 N/m 2 )(1.0 N/ m) πσ 2 c (π) 13.5 x10 6 N/m 2 ( ) x 10-4 m 0.79 mm (0.031 in.) 8.16 This problem asks us to determine the stress level at which an aircraft component will fracture for a given fracture toughness (26 MPa m ) and maximum internal crack length (6.0 mm), given that fracture occurs for the same component using the same alloy at one stress level and another internal crack length. It first becomes necessary to solve for the parameter Y for the conditions under which fracture occurred using Equation (8.5). Therefore, Y K Ic σ πa 26 MPa m (112 MPa) (π) 8.6 x 10 3 m Now we will solve for σ c using Equation (8.6) as

14 σ c K Ic Y π a 26 MPa m 6 x 10 3 m (2.0) (π) MPa (19,300 psi) 8.23 The plot of impact energy versus temperature is shown below. (b) The average of the maximum and minimum impact energies from the data is Average 76 J + 2J 2 39 J As indicated on the plot by the one set of dashed lines, the ductile-to-brittle transition temperature according to this criterion is about 10 C. (c) Also, as noted on the plot by the other set of dashed lines, the ductile-to-brittle transition temperature for an impact energy of 20 J is about -2 C (a) The fatigue data for this alloy are plotted below.

15 (b) As indicated by one set of dashed lines on the plot, the fatigue strength at 4 x 10 6 cycles [log (4 x 10 6 ) 6.6] is about 100 MPa. (c) As noted by the other set of dashed lines, the fatigue life for 120 MPa is about 6 x 10 5 cycles (i.e., the log of the lifetime is about 5.8) This problem asks that we determine the total elongation of a low carbon-nickel alloy that is exposed to a tensile stress of 70 MPa (10,000 psi) at 427 C for 10,000 h; the instantaneous and primary creep elongations are 1.3 mm (0.05 in.). From the 427 C line in Figure 8.29, the steady state creep rate,?ε s, is about %/1000 h (or 3.5 x 10-5 %/h) at 70 MPa. The steady state creep strain, ε s, therefore, is just the product of?ε s and time as ε s?ε s x (time) ( 3.5 x 10-5 %/h)(10,000 h) 0.35 % 3.5 x10-3 Strain and elongation are related as in Equation (6.2); solving for the steady state elongation, Δl s, leads to Δl s l o ε s (1015 mm) ( 3.5 x 10-3 ) 3.6 mm (0.14 in.)

16 Finally, the total elongation is just the sum of this Δl s and the total of both instantaneous and primary creep elongations [i.e., 1.3 mm (0.05 in.)]. Therefore, the total elongation is 4.9 mm (0.19 in.). 8.D4W This problem asks that we consider a steel plate having a through-thickness edge crack, and to determine if fracture will occur given the following: W 40 mm, B 6.0 mm, K Ic 60 MPa m (54.6 ksi in.), σ y 1400 MPa, σ 200 MPa, and a 16 mm. The first thing to do is determine whether conditions of plane strain exist. From Equation (8.14W), 2 K Ic 2.5 σ y 2 60 MPa m m 4.6 mm (0.19 in.) 1400 MPa Inasmuch as the plate thickness is 6 mm (which is greater than 4.6 mm), the situation is a plane strain one. Next, we must determine the a/w ratio, which is just 16 mm/40 mm From this ratio and using Figure 8.7aW, Y At this point it becomes necessary to determine the value of the Yσ π a product; if it is greater than K Ic then fracture will occur. Thus Yσ πa (2.12)(200 MPa) (π)( 16 x10 3 m) 95.0 MPa m ( 86.5 ksi in. ) Therefore, fracture will occur since this value (95.0 MPa m) is greater than the K Ic for the steel (60 MPa m).

17 9.5 This problem asks that we cite the phase or phases present for several alloys at specified temperatures. (a) For an alloy composed of 15 wt% Sn-85 wt% Pb and at 100 C, from Figure 9.7, α and β phases are present, and C α 5 wt% Sn-95 wt% Pb C β 98 wt% Sn-2 wt% Pb (b) For an alloy composed of 25 wt% Pb-75 wt% Mg and at 425 C, from Figure 9.18, only the α phase is present; its composition is 25 wt% Pb-75 wt% Mg. (c) For an alloy composed of 85 wt% Ag-15 wt% Cu and at 800 C, from Figure 9.6, β and liquid phases are present, and C β 92 wt% Ag-8 wt% Cu C L 77 wt% Ag-23 wt% Cu (d) For an alloy composed of 55 wt% Zn-45 wt% Cu and at 600 C, from Figure 9.17, β and γ phases are present, and C β 51 wt% Zn-49 wt% Cu C γ 58 wt% Zn-42 wt% Cu (e) For an alloy composed of 1.25 kg Sn and 14 kg Pb and at 200 C, we must first determine the Sn and Pb concentrations, as C Sn 1.25 kg wt% 1.25 kg + 14 kg C Pb 14 kg wt% 1.25 kg + 14 kg From Figure 9.7, only the α phase is present; its composition is 8.2 wt% Sn-91.8 wt% Pb. (f) For an alloy composed of 7.6 lb m Cu and lb m Zn and at 600 C, we must first determine the Cu and Zn concentrations, as

18 C Cu 7.6 lb m 7.6 lb m lb m wt% C Zn lb m 7.6 lb m lb m wt% From Figure 9.17, only the L phase is present; its composition is 95.0 wt% Zn-5.0 wt% Cu (g) For an alloy composed of 21.7 mol Mg and 35.4 mol Pb and at 350 C, it is first necessary to determine the Mg and Pb concentrations, which we will do in weight percent as follows: ' m Pb n mpb A Pb (35.4 mol)(207.2 g/mol) 7335 g ' m Mg n mmg A Mg (21.7 mol)(24.3 g/mol) 527 g C Pb 7335 g wt% 7335 g g C Mg 100 wt% 93 wt% 7 wt% From Figure 9.18, L and Mg 2 Pb phases are present, and C L 94 wt% Pb 6wt%Mg C Mg2 Pb 81 wt% Pb 19 wt% Mg (h) For an alloy composed of 4.2 mol Cu and 1.1 mol Ag and at 900 C, it is first necessary to determine the Cu and Ag concentrations, which we will do in weight percent as follows:

19 ' m Cu n mcu A Cu (4.2 mol)(63.55 g/mol) g ' m Ag n mag A Ag (1.1 mol)( g/mol) g C Cu g g g wt% C Ag g wt% g g From Figure 9.6, α and liquid phases are present; and C α 8 wt% Ag-92 w% Cu C L 45 wt% Ag-55 wt% Cu 9.7 This problem asks that we determine the phase mass fractions for the alloys and temperatures in Problem 9.5. (a) W α C β C o C β C α W β C o C α C β C α (b) W α 1.0 (c) W β C o C L C β C L W L C β C o C β C L

20 (d) W β C γ C o C C γ β W γ C o C β C γ C β (e) W α 1.0 (f) W L 1.0 (g) W Mg2 Pb C L C o C L C Mg2 Pb W L C o C L C Mg2 Pb C Mg2 Pb (h) W α C L C o C L C α W L C o C α C L C α (a) This part of the problem calls for us to cite the temperature to which a 90 wt% Pb-10 wt% Sn alloy must be heated in order to have 50% liquid. Probably the easiest way to solve this problem is by trial and error--that is, moving vertically at the given composition, through the α + L region until the tie-line lengths on both sides of the given composition are the same (Figure 9.7). This occurs at approximately 300 C (570 F). (b) We can also produce a 50% liquid solution at 250 C, by adding Sn to the alloy. At 250 C and within the α + L phase region

21 C α 13 wt% Sn-87 wt% Pb C L 39 wt% Sn-61 wt% Pb Let C o be the new alloy composition to give W α W L 0.5. Then, W α 0.5 C L C o 39 C o C L C α And solving for C o gives 26 wt% Sn. Now, let m Sn be the mass of Sn added to the alloy to achieve this new composition. The amount of Sn in the original alloy is (0.10)(1.5 kg) 0.15 kg Then, using a modified form of Equation (4.3) 0.15 kg + m Sn 1.5 kg + m Sn And, solving for m Sn yields m Sn kg This problem asks if a noncold-worked Cu-Ni solid solution alloy is possible having a minimum tensile strength of 380 MPa (55,000 psi) and also a ductility of at least 45%EL. From Figure 9.5a, a tensile strength greater than 380 MPa is possible for compositions between about 34 and 87 wt% Ni. On the other hand, according to Figure 9.5b, ductilities greater than 45%EL exist for compositions less than about 7 wt% and greater than about 71 wt% Ni. Therefore, the stipulated criteria are met for all compositions between 71 and 87 wt% Ni Upon cooling a 50 wt% Ni-50 wt% Cu alloy from 1400 C and utilizing Figure 9.2a: (a) The first solid phase forms at the temperature at which a vertical line at this composition intersects the L-(α + L) phase boundary--i.e., at about 1320 C; (b) The composition of this solid phase corresponds to the intersection with the L-(α + L) phase boundary, of a tie line constructed across the α + L phase region at 1320 C--i.e., C α 62 wt% Ni-38 wt% Cu;

22 (c) Complete solidification of the alloy occurs at the intersection of this same vertical line at 50 wt% Ni with the (α + L)-α phase boundary--i.e., at about 1270 C; (d) The composition of the last liquid phase remaining prior to complete solidification corresponds to the intersection with the L-(α + L) boundary, of the tie line constructed across the α + L phase region at 1270 C--i.e., C L is about 37 wt% Ni-63 wt% Cu Yes, it is possible to have a Cu-Ag alloy of composition 20 wt% Ag-80 wt% Cu which consists of mass fractions W α 0.80 and W L Using the appropriate phase diagram, Figure 9.6, by trial and error with a ruler, the tie-line segments within the α + L phase region are proportioned such that W α 0.8 C L C L C o C α for C o 20 wt% Ag. This occurs at about 800 C This problem asks if it is possible to have a Mg-Pb alloy for which the mass fractions of primary α and total α are 0.60 and 0.85, respectively, at 460 C. In order to make this determination we need to set up the appropriate lever rule expression for each of these quantities. From Figure 9.18 and at 460 C, C α 41 wt% Pb, C Mg2 Pb 81 wt% Pb, and C eutectic 67 wt% Pb. For primary α W α' C eutectic C o C eutectic C α 67 C o Solving for C o gives C o 51.4 wt% Pb. Now the analogous expression for total α W α C Mg2 Pb C o C Mg2 Pb C α 81 C o And this value of C o is 47 wt% Pb. Therefore, since these two C o values are different, this alloy is not possible.

23 9.45 Below is shown the phase diagram for these two A and B metals This problem asks us to consider various aspects of 1.0 kg of austenite containing 1.15 wt% C that is cooled to below the eutectoid. (a) The proeutectoid phase will be Fe C since 1.15 wt% C is greater than the eutectoid 3 (0.76 wt% C). (b) For this portion of the problem, we are asked to determine how much total ferrite and cementite form. Application of the appropriate lever rule expression yields W α C Fe3 C C o C Fe3 C C α which, when multiplied by the total mass of the alloy (1.0 kg), gives 0.83 kg of total ferrite. Similarly, for total cementite, W Fe3 C C o C α C Fe3 C C α

24 And the mass of total cementite that forms is (0.17)(1.0 kg) 0.17 kg. (c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite) form. Applying Equation (9.20), in which C 1 ' 1.15 wt% C W p 6.70 C 1 ' which corresponds to a mass of 0.93 kg. Likewise, from Equation (9.21) W Fe3 C' C 1 ' which is equivalent to 0.07 kg of the total 1 kg mass. (d) Schematically, the microstructure would appear as: 9.59 In this problem we are asked to consider 2.0 kg of a 99.6 wt% Fe-0.4 wt% C alloy that is cooled to a temperature below the eutectoid. (a) Equation (9.19) must be used in computing the amount of proeutectoid ferrite that forms. Thus, W α' 0.76 C o ' Or, (0.49)(2.0 kg) 0.99 kg of proeutectoid ferrite forms.

25 (b) In order to determine the amount of eutectoid ferrite, it first becomes necessary to compute the amount of total ferrite using the lever rule applied entirely across the α + Fe 3 C phase field, as W α C Fe3 C C o ' C Fe3 C C α which corresponds to (0.94)(2.0 kg) 1.89 kg. Now, the amount of eutectoid ferrite is just the difference between total and proeutectoid ferrites, or 1.89 kg kg 0.90 kg (c) With regard to the amount of cementite that forms, again application of the lever rule across the entirety of the α + Fe 3 C phase field, leads to W Fe3 C C o ' C α C Fe3 C C α which amounts to (0.06)(2 kg) 0.11 kg cementite in the alloy For this problem, we are given, for the austenite-to-pearlite transformation, two values of y and two values of the corresponding times, and are asked to determine the time required for 95% of the austenite to transform to pearlite. The first thing necessary is to set up two expressions of the form of Equation (10.1), and then to solve simultaneously for the values of n and k. In order to expedite this process, we will rearrange and do some algebraic manipulation of Equation (10.1). First of all, we rearrange as follows: 1 y exp ( kt n ) Now taking natural logarithms ln (1 y) kt n

26 Or ln (1 y) kt n which may also be expressed as 1 ln kt n 1 y Now taking natural logarithms again, leads to 1 ln ln 1 y ln k + nlnt which is the form of the equation that we will now use. The two equations are thus 1 ln ln ln k + n ln(280 s) 1 ln ln ln k + n ln(425 s) Solving these two expressions simultaneously for n and k yields n and k x Now it becomes necessary to solve for the value of t at which y Algebraic manipulation of Equation (10.1) leads to an expression in which t is the dependent parameter as t ln (1 y) k 1/n 1/3.385 ln (1 0.95) x s

27 10.7 This problem asks us to consider the percent recrystallized versus logarithm of time curves for copper shown in Figure (a) The rates at the different temperatures are determined using Equation (10.2), which rates are tabulated below: Temperature ( C) Rate (min) x x x x x 10-5 (b) These data are plotted below. The activation energy, Q, is related to the slope of the line drawn through the data points as

28 Q Slope (R) where R is the gas constant. The slope of this line is x 10 4 K, and thus Q ( x 10 4 K)(8.31 J/mol - K) 93,600 J/mol (c) At room temperature (20 C), 1/T 3.41 x 10-3 K -1. Extrapolation of the data in the plot to this 1/T value gives ln (rate) 12.8 which leads to rate exp ( 12.8) 2.76 x 10-6 (min) -1 But since rate 1 t 0.5 then t rate x 10 6 (min) x 10 5 min 250 days (a) Superheating and supercooling correspond, respectively, to heating or cooling above or below a phase transition temperature without the occurrence of the transformation. (b) Superheating and supercooling occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. The driving force is enhanced during superheating or supercooling.

29 10.14 This problem asks us to determine the nature of the final microstructure of an ironcarbon alloy of eutectoid composition, that has been subjected to various isothermal heat treatments. Figure is used in these determinations. (a) 50% coarse pearlite and 50% martensite (b) 100% spheroidite (c) 50% fine pearlite, 25% bainite, and 25% martensite (d) 100% martensite (e) 40% bainite and 60% martensite (f) 100% bainite (g) 100% fine pearlite (h) 100% tempered martensite Below is shown an isothermal transformation diagram for a 1.13 wt% C iron-carbon alloy, with time-temperature paths that will produce (a) 6.2% proeutectoid cementite and 93.8% coarse pearlite; (b) 50% fine pearlite and 50% bainite; (c) 100% martensite; and (d) 100% tempered martensite.

30 10.20 Below is shown a continuous cooling transformation diagram for a 0.35 wt% C ironcarbon alloy, with continuous cooling paths that will produce (a) fine pearlite and proeutectoid ferrite; (b) martensite; (c) martensite and proeutectoid ferrite; (d) coarse pearlite and proeutectoid ferrite; and (e) martensite, fine pearlite, and proeutectoid ferrite.

31 10.25 This problem asks that we briefly describe the simplest continuous cooling heat treatment procedure that would be used in converting a 4340 steel from one microstructure to another. Solutions to this problem require the use of Figure (a) In order to convert from (martensite + ferrite + bainite) to (martensite + ferrite + pearlite + bainite) it is necessary to heat above about 720 C, allow complete austenitization, then cool to room temperature at a rate between 0.02 and C/s. (b) To convert from (martensite + ferrite + bainite) to spheroidite the alloy must be heated to about 700 C for several hours. (c) In order to convert from (martensite + bainite + ferrite) to tempered martensite it is necessary to heat to above about 720 C, allow complete austenitization, then cool to room temperature at a rate greater than 8.3 C/s, and finally isothermally heat treat the alloy at a temperature between about 400 and 550 C (Figure 10.25) for about one hour This problem asks us to rank four iron-carbon alloys of specified composition and microstructure according to tensile strength. This ranking is as follows: 0.6 wt% C, fine pearlite 0.6 wt% C, coarse pearlite

32 0.25 wt% C, coarse pearlite 0.25 wt% C, spheroidite The 0.25 wt% C, coarse pearlite is stronger than the 0.25 wt% C, spheroidite since coarse pearlite is stronger than spheroidite; the composition of the alloys is the same. The 0.6 wt% C, coarse pearlite is stronger than the 0.25 wt% C, coarse pearlite, since increasing the carbon content increases the strength. Finally, the 0.6 wt% C, fine pearlite is stronger than the 0.6 wt% C, coarse pearlite inasmuch as the strength of fine pearlite is greater than coarse pearlite because of the many more ferrite-cementite phase boundaries in fine pearlite This problem asks for Rockwell hardness values for specimens of an iron-carbon alloy of eutectoid composition that have been subjected to some of the heat treatments described in Problem (b) The microstructural product of this heat treatment is 100% spheroidite. According to Figure 10.21(a) the hardness of a 0.76 wt% C alloy with spheroidite is about 87 HRB (180 HB). (d) The microstructural product of this heat treatment is 100% martensite. According to Figure 10.23, the hardness of a 0.76 wt% C alloy consisting of martensite is about 64 HRC (690 HB). (f) The microstructural product of this heat treatment is 100% bainite. From Figure 10.22, the hardness of a 0.76 wt% C alloy consisting of bainite is about 385 HB. And, conversion from Brinell to Rockwell hardness using Figure 6.18 leads to a hardness of 36 HRC. (g) The microstructural product of this heat treatment is 100% fine pearlite. According to Figure 10.21(a), the hardness of a 0.76 wt% C alloy consisting of fine pearlite is about 27 HRC (270 HB). (h) The microstructural product of this heat treatment is 100% tempered martensite. According to Figure 10.26, the hardness of a water-quenched eutectoid alloy that was tempered at 315 C for one hour is about 56 HRC (560 HB). 10.D2 This problem asks if it is possible to produce an iron-carbon alloy that has a minimum tensile strength of 620 MPa (90,000 psi) and a minimum ductility of 50%RA. If such an alloy is possible, its composition and microstructure are to be stipulated. From Equation (6.20a), this tensile strength corresponds to a Brinell hardness of

33 HB TS(MPa) 620 MPa According to Figures 10.21(a) and (b), the following is a tabulation of the composition ranges for fine and coarse pearlites and spheroidite which meet the stipulated criteria. Compositions for Compositions for Microstructure HB 180 %RA 50% Fine pearlite > 0.38 %C < 0.36 %C Coarse pearlite > 0.47 %C < 0.42 %C Spheroidite > 0.80 %C %C Therefore, only spheroidite has a composition range overlap for both of the hardness and ductility restrictions; the spheroidite would necessarily have to have a carbon content greater than 0.80 wt% C The alloying elements in tool steels (e.g., Cr, V, W, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds The principal difference between wrought and cast alloys is as follows: wrought alloys are ductile enough so as to be hot- or cold-worked during fabrication, whereas cast alloys are brittle to the degree that shaping by deformation is not possible and they must be fabricated by casting Rivets of a 2017 aluminum alloy must be refrigerated before they are used because, after being solution heat treated, they precipitation harden at room temperature. Once precipitation hardened, they are too strong and brittle to be driven. 11.D1 This problem calls for us to select, from a list, the best alloy for each of several applications and then to justify each choice. (a) Gray cast iron would be the best choice for a milling machine base because it effectively absorbs vibrations and is inexpensive. (b) Stainless steel would be the best choice for the walls of a steam boiler because it is corrosion resistant to the steam and condensate. (c) Titanium alloys are the best choice for high-speed aircraft because they are light weight, strong, and easily fabricated. However, one drawback is their cost.

34 (d) A tool steel would be the best choice for a drill bit because it is very hard and wear resistant, and, thus, will retain a sharp cutting edge. (e) For a cryogenic (low-temperature) container, an aluminum alloy would be the best choice; aluminum alloys have the FCC crystal structure, and therefore, are ductile down to very low temperatures. (f) As a pyrotechnic in flares and fireworks, magnesium is the best choice because it ignites easily and burns readily in air. (g) Platinum is the best choice for high-temperature furnace elements to be used in oxidizing atmospheres because it is very ductile, has a relatively high melting temperature, and is highly resistant to oxidation. 11.D3 The first part of this question asks for a description of the shape memory phenomenon. A part having some shape and that is fabricated from a metal alloy that displays this phenomenon is plastically deformed. It can be made to return to its original shape by heating to an elevated temperature. Thus, the material has a shape memory, or "remembers" its previous shape. Next we are asked to explain the mechanism for this phenomenon. A shape memory alloy is polymorphic (Section 3.6)--that is, it can exist having two crystal structures. One is body-centered cubic structure (termed an austenite phase) that exists at elevated temperatures; upon cooling, and at some temperature above the ambient, it transforms to a martensitic structure. Furthermore, this martensitic phase is highly twinned. Upon application of a stress to this low-temperature martensitic phase, plastic deformation is accomplished by the migration of twin boundaries to some preferred orientation. Once the stress is removed, the deformed shape will be retained at this temperature. When this deformed martensite is subsequently heated to above the phase transformation temperature, the alloy reverts back to the BCC phase, and assumes the original shape. The procedure may then be repeated. One material that exhibits this behavior is a nickel-titanium alloy. Furthermore, the desired "memory" shape may is established by forming the material above its phase transition temperature. Several applications for alloys displaying this effect are eyeglass frames, shrink-tofit pipe couplers, tooth-straightening braces, collapsible antennas, greenhouse window openers, antiscald control valves on showers, women's foundations, and fire sprinkler valves.

35 11.D7 This question asks for us to decide whether or not it would be advisable to hot-work or cold-work several metals and alloys. Tin would almost always be hot-worked. Even deformation at room temperature would be considered hot-working inasmuch as its recrystallization temperature is below room temperature (Table 7.2). Tungsten is hard and strong at room temperature, has a high recrystallization temperature, and experiences oxidation at elevated temperatures. Cold-working is difficult because of its strength, and hot-working is not practical because of oxidation problems. Most tungsten articles are fabricated by powder metallurgy, or by using cold-working followed by annealing cycles. Most aluminum alloys may be cold-worked since they are ductile and have relatively low yield strengths. Magnesium alloys are normally hot-worked inasmuch as they are quite brittle at room temperature. Also, magnesium alloys have relatively low recrystallization temperatures. A 4140 steel could be cold-worked in an over-tempered state which leaves it soft and relatively ductile, after which quenching and tempering heat treatments may be employed to strengthen and harden it. This steel would probably have a relatively high recrystallization temperature, and hot-working may cause oxidation. 11.D9 (a) This problem calls for us to decide which of 8660, 8640, 8630, and 8620 alloys may be fabricated into a cylindrical piece 57 mm (2-1/4 in.) in diameter which, when quenched in mildly agitated water, will produce a minimum hardness of 45 HRC throughout the entire piece. The center of the steel cylinder will cool the slowest and therefore will be the softest. In moderately agitated water the equivalent distance from the quenched end for a 57 mm diameter bar for the center position is about 11 mm (7/16 in.) [Figure 11.16(a)]. The hardnesses at this position for the alloys cited (Figure 11.14) are given below. Alloy Center Hardness (HRC)

36 Therefore, only 8660 and 8640 alloys will have a minimum of 45 HRC at the center, and therefore, throughout the entire cylinder. (b) This part of the problem asks us to do the same thing for moderately agitated oil. In moderately agitated oil the equivalent distance from the quenched end for a 57 mm diameter bar at the center position is about 17.5 mm (11.16 in.) [Figure 11.16(b)]. The hardnesses at this position for the alloys cited (Figure 11.14) are given below. Alloy Center Hardness (HRC) Therefore, only the 8660 alloy will have a minimum of 45 HRC at the center, and therefore, throughout the entire cylinder.